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2 years ago

The form of energy that can move from place to place across the universe is . On Earth, the main source of this energy is .

2 answers:

5 0

The form of energy that can move from place to place across the universe is light energy. On earth, the main source of this energy is Sun. Most of the light energy comes from the sun because it is the primary source of all the energies. The food, fossil fuels, movement of winds, etc all exists due to Sun. Without sun, there won't be any light energy on the earth. In all the processes which occur on earth has a direct or indirect involvement of light energy which comes from sun.

iren2701 [21]2 years ago

5 0

Answer: Light energy; sunlight

Explanation:

The form of energy that can move from one place to another is light energy. The energy of light depends on the intensity.

Sunlight is the main source of energy on earth, this gives rise to other forms of energy available on earth.

Plants consume light energy to make food which is a source of energy for all other live forms on earth.

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PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! A IS NOT THE CORRECT ANSWER

patriot [66]

Answer: A

Explanation:

NOTES:

d = 650 meters

t = 10 seconds

**********************************

v = d/t

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= 65 meters/second

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3 years ago

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sergey [27]

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3 years ago

You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli

FinnZ [79.3K]

Answer:

Explanation:

A general wave function is given by:

$f(x,t)=Acos(kx-\omega t)$

A: amplitude of the wave = 0.075m

k: wave number

w: angular frequency

a) You use the following expressions for the calculation of k, w, T and λ:

$\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}$

$k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}$

$T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m$

b) Hence, the wave function is:

$f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)$

c) for x=3m you have:

$f(3,t)=0.075cos(1.047*3-12.56t)$

d) the speed of the medium:

$\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)$

you can see the velocity of the medium for example for x = 0:

$v=\frac{df}{dt}=13.15cos(12.56t)$

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3 years ago

What is a characteristic of a gas?

DerKrebs [107]

Answer:

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Explanation:

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2 years ago

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Answer:

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